Answer:
The probability that she takes out a red marble the first time and a green marble the second
P(E₁ ∩ E₂ ) = P(E₁) P(E₂)
![= \frac{5 X 5}{20} = \frac{25}{20} = \frac{5}{4} ways](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B5%20X%205%7D%7B20%7D%20%3D%20%5Cfrac%7B25%7D%7B20%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D%20ways)
Step-by-step explanation:
<u><em>Step(i)</em></u>:-
Given a bag of marbles contains 5 red marbles, 5 green marbles, 5 yellow marbles and 5 blue marbles
<em>Total balls n(S) = 5red +5green +5yellow + 5blue = 20</em>
<em>Total numbers cases </em>
<em> n(S) = 20C₁ = 20 ways</em>
<u><em>Step(ii):</em></u><em>-</em>
<em>Let E₁ be the event of drawing one red marble from 5 red balls</em>
<em> n(E₁) = 5C₁ </em>
Let E₂ be the event of drawing one green marble from 5 green balls
n(E₂) = 5C₁
<em>The two events are independent events</em>
P(E₁ ∩ E₂ ) = P(E₁) P(E₂)
=![= \frac{5 X 5}{20} = \frac{25}{20} = \frac{5}{4} ways](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B5%20X%205%7D%7B20%7D%20%3D%20%5Cfrac%7B25%7D%7B20%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D%20ways)
<u><em>Final answer:</em></u>-
<em> The probability that she takes out a red marble the first time and a green marble the second</em>
P(E₁ ∩ E₂ ) = P(E₁) P(E₂)
=![= \frac{5 X 5}{20} = \frac{25}{20} = \frac{5}{4} ways](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B5%20X%205%7D%7B20%7D%20%3D%20%5Cfrac%7B25%7D%7B20%7D%20%3D%20%5Cfrac%7B5%7D%7B4%7D%20ways)